__Activation energy__

__Activation energy__

- The minimum quantitiy of external energy required for the conversion of reactant into product or to produce an unstable intermediate is called activation energy. It is E
- Rate of reaction is inversely proportional to the activation energy.
- Therefore, greater value of activation energy leads to lower rate of reaction and increased influence of temperature change on the rate constant.

** PROBLEM**. The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate E

_{Solution}.

** SOLUTION. **It is given that T

_{1}= 298 K

∴T2 = (298 + 10) K

= 308 K

We also know that the rate of the reaction doubles when temperature is increased by 10°.

Therefore, let us take the value of k_{1} = k and that of k_{2} = 2k

Also, R = 8.314 J K ^{– 1} mol ^{– 1}

Now, substituting these values in the equation:

Log k_{2}/k_{1} = E_{a}/2.303R [T_{2}-T_{1}/T_{1}T_{2}]

Log 2k/k = E_{a}/2.303 X 8.314 [10/298 X 308]

E_{a}= 2.303 X 8.314 X 298 X 308 X log2/10

= 52897.78 J mol^{ – 1}

= 52.9 kJ mol^{ – 1}

__Arrhenius equation__

__Arrhenius equation__

- The formula used to calculate the energy of activation and justify the effect of temperature on rate of reaction is called Arrhenius Equation.
- It is given by the formula,

K = A e^{-Ea/RT}

Where,

k = Rate constant

A= Frequency factor

e = mathematical quantity

E_{a}= activation energy

R = gas constant

T = kelvin temperature

ln K = ln A – Ea/(2.303RT)

Equation of a straight line with slope = –*E _{a}* /

*R*.

- When E
_{a}= 0 , Temperature = Infinity

K = Ae^{0} = A

e^{-Ea/RT }=Boltzmann factor.

- For a chemical reaction the rate constant gets doubled for a rise of 10° temperature. This is because according to Arrhenius Equation,

K = Ae^{-Ea/RT}

Taking log on both sides of the equation

Ln k = ln A – E_{a}/RT

Comparing with the equation of a straight line

y= mx+c,

[m= slope of the line

c= y-intercept]

So we have:

y = ln k

x = 1/T

m = -E_{a} / R

c = ln A

Plotting k Vs (1/*T*)

** PROBLEM**. Find the activation energy (in kJ/mol) of the reaction if the rate constant at 600K is 3.4 M

^{-1}s

^{-1}and 31.0 at 750K.

** SOLUTION**. Ln k = ln A – E

_{a}/RT

To find E_{a}, subtract ln A from both sides and multiply by -RT.

E_{a} = (ln A – ln k)RT

__Collision theory__

__Collision theory__

- It states that:
- According to collision theory the molecules collides with great kinetic energy in order to bring about a chemical reaction.

*The molecules of the reacting species collide through the space in a rectilinear motion.*

*Rate of a chemical reaction is proportional to the number of collisions between the molecules of the reacting species.*

*The molecules must be properly oriented.*

- Rate of successful collisions ∝ Fraction of successful collisions X Overall collision frequency.
- The number of collisions per second per unit volume of the molecules in a chemical reaction is called collision frequency (Z).

Let A+B –> C + D

Rate = Z_{AB}e^{-Ea/RT}

Here Z_{AB }= collision frequency of A and B.

- In many reactions Rate = P Z
_{AB}e^{-Ea/RT}

Where p= steric factor which takes into account the proper orientation of the molecules participating in a chemical reaction.

**PROBLEM. **The activation energy for the reaction 2HI_{(g)} → H_{2} + I_{2(g)} is 209.5 kJ mol^{-1} at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

**SOLUTION. ** *E*_{a} = 209.5 kJ mol^{ – 1} = 209500 J mol^{ – 1}

*T* = 581 K

*R* = 8.314 JK^{ – 1} mol^{ – 1}

Fraction of molecules of reactants having energy equal to or greater than activation energy is as follows:

x = e^{-Ea/RT}

ln x = -Ea/RT

log x = Ea/2.303RT

log x = 209500 J mol^{-1}/2.303 X 8.314 J K^{-1 }mol^{-1} X 581 = 18.8323

x = antilog (18.323)

= antilog 19.1977

= 1.471 X 10^{-19}